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# Is |x + y| > |x - y| ?

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Intern
Joined: 06 Feb 2013
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Is |x + y| > |x - y| ?  [#permalink]

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Updated on: 20 Dec 2017, 20:09
2
17
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Difficulty:

75% (hard)

Question Stats:

59% (02:19) correct 41% (02:24) wrong based on 369 sessions

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Is |x + y| > |x - y| ?

(1) |x| > |y|
(2) |x - y| < |x|

Originally posted by josemarioamaya on 25 Sep 2013, 13:07.
Last edited by Bunuel on 20 Dec 2017, 20:09, edited 2 times in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 50623
Re: Is |x + y| > |x - y| ?  [#permalink]

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25 Sep 2013, 13:18
11
8
Is |x + y| > |x - y| ?

Is $$|x+y| > |x-y|$$? --> square it: is $$x^2+2xy+y^2>x^2-2xy+y^2$$ --> is $$xy>0$$?

(1) |x| > |y|. We cannot get from this whether x and y have the same sign. Not sufficient.

(2) |x - y| < |x| --> square again: $$x^2-2xy+y^2<x^2$$ --> $$xy>\frac{y^2}{2}$$ --> since $$y^2\geq{0}$$ (the square of any number is more than or equal to 0), then we have that $$xy>\frac{y^2}{2}\geq{0}$$. Sufficient.

Hope it's clear.
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Is |x + y| > |x - y| ?  [#permalink]

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29 Jul 2018, 18:47
1
Mo2men wrote:
GMATGuruNY
Can you please show how I can open the modulus in statement 2 instead of squaring both sides .

Thanks

Statement 2: |x-y| < |x|

Case 1: x>0 and x≥y
In this case, |x-y| = x-y and |x| = x.
Substituting the blue equations into |x-y| < x, we get:
x-y < x
-y < 0
y > 0
Implication:
If x>0 and x≥y, then y>0.
Thus:
0<y≤x

Case 2: x>0 and x<y
In this case, |x-y| = -x+y and |x| = x.
Substituting the blue equations into |x-y| < x, we get:
-x+y < x
y < 2x
Implication:
If x>0 and x<y, then y<2x.
Thus:
0<x<y<2x

Case 3: x<0 and x≥y
In this case, |x-y| = x-y and |x| = -x.
Substituting the blue equations into |x-y| < x, we get:
x-y < -x
2x < y
Implication:
If x<0 and x≥y, then y>2x.
Thus:
2x<y≤x<0

Case 4: x<0 and x<y
In this case, |x-y| = -x+y and |x| = -x.
Substituting the blue equations into |x-y| < x, we get:
-x+y < -x
y < 0
Implication:
If x<0 and x<y, then y<0.
Thus:
x<y<0

In all four cases, x and y have the same sign, so xy>0.

I do not recommend this approach.
Squaring the inequality seems easier and faster.
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##### General Discussion
Intern
Joined: 29 Aug 2012
Posts: 29
WE: General Management (Consulting)
Re: Is |x + y| > |x - y| ?  [#permalink]

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26 Sep 2013, 08:16
Bunuel wrote:
Is |x + y| > |x - y| ?

Is $$|x+y| > |x-y|$$? --> square it: is $$x^2+2xy+y^2>x^2-2xy+y^2$$ --> is $$xy>0$$?

(1) |x| > |y|. We cannot get from this whether x and y have the same sign. Not sufficient.

(2) |x - y| < |x| --> square again: $$x^2-2xy+y^2<x^2$$ --> $$xy>\frac{y^2}{2}$$ --> since $$y^2\geq{0}$$ (the square of any number is more than or equal to 0), then we have that $$xy>\frac{y^2}{2}\geq{0}$$. Sufficient.

Hope it's clear.

Sorry Bunuel,

but i don't understand the concept behind the solution...

I understand that an absolute value is always positive. But how can you just square it¿? It supposed to be the root square of the square, isn't it¿??
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Posts: 50623
Re: Is |x + y| > |x - y| ?  [#permalink]

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26 Sep 2013, 08:19
jacg20 wrote:
Bunuel wrote:
Is |x + y| > |x - y| ?

Is $$|x+y| > |x-y|$$? --> square it: is $$x^2+2xy+y^2>x^2-2xy+y^2$$ --> is $$xy>0$$?

(1) |x| > |y|. We cannot get from this whether x and y have the same sign. Not sufficient.

(2) |x - y| < |x| --> square again: $$x^2-2xy+y^2<x^2$$ --> $$xy>\frac{y^2}{2}$$ --> since $$y^2\geq{0}$$ (the square of any number is more than or equal to 0), then we have that $$xy>\frac{y^2}{2}\geq{0}$$. Sufficient.

Hope it's clear.

Sorry Bunuel,

but i don't understand the concept behind the solution...

I understand that an absolute value is always positive. But how can you just square it¿? It supposed to be the root square of the square, isn't it¿??

If both sides of an inequality are non negative, then we can square it.

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Re: Is |x + y| > |x - y| ?  [#permalink]

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09 May 2014, 14:25
1
I solved in the following way

First square both sides

We get 'Is xy>0" , or in proper english do 'x' and 'y' have the same sign?

Statement 1: Obviously insufficient
Statement 2: Square both sides again ---> x^2 > x^2 - 2xy + y^2

y^2 - 2xy < 0

y (y-2x)<0

So two cases:

If y>0, then y-2x<0 and 2x>y>0. Answer is YES
If y<0, then y-2x>0 and 2x<y<0. Answer is YES again

Sufficient

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Re: Is |x + y| > |x - y| ?  [#permalink]

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04 Nov 2014, 10:05
1
Basically the question is asking if x and y have the same sign.

from statement 1: no info on signs.---NSF

for statement 2: I did the following:

The distance between x and y on the number line can be less than the distance of 0 and x only when x and y have the same sign.

---------y----x---------0--------------
or
----------0---------x---y-------------

if
------x-------0--------y---- or ---------y--------0------------x----------
then the statement-2 is not true.

Therefore x and y should have the same sign.
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Posts: 566
Re: Is |x + y| > |x - y| ?  [#permalink]

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25 Jul 2018, 08:55
Statement 1 is insufficient when you test for positive & negative numbers. For e.g. (8,7) & (-8,7)

Statement 2 is sufficient hence Option B is the correct answer.
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Is |x + y| > |x - y| ?  [#permalink]

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27 Jul 2018, 03:29
Bunuel wrote:
Is |x + y| > |x - y| ?

Is $$|x+y| > |x-y|$$? --> square it: is $$x^2+2xy+y^2>x^2-2xy+y^2$$ --> is $$xy>0$$?

(1) |x| > |y|. We cannot get from this whether x and y have the same sign. Not sufficient.

(2) |x - y| < |x| --> square again: $$x^2-2xy+y^2<x^2$$ --> $$xy>\frac{y^2}{2}$$ --> since $$y^2\geq{0}$$ (the square of any number is more than or equal to 0), then we have that $$xy>\frac{y^2}{2}\geq{0}$$. Sufficient.

Hope it's clear.

GMATGuruNY
Can you please show how I can open the modulus in statement 2 instead of squaring both sides .

Thanks
Is |x + y| > |x - y| ? &nbs [#permalink] 27 Jul 2018, 03:29
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